Answer
The domain of $tan^{-1}~(1+e^{-t^2})$ is $(-\infty, \infty)$ and this function is continuous at every number in its domain.
Work Step by Step
$N(r) = tan^{-1}(1+e^{-t^2})$
The domain of $f(t) = tan^{-1}~t~~$ is all real numbers.
According to Theorem 7, $~~f(t) = tan^{-1}~t~~$ is continuous for all $t$
The function $~~g(t) = 1+e^{-t^2}~~$ is continuous for all values of $t$
According to Theorem 9, $f(g(t)) = tan^{-1}~(1+e^{-t^2})$ is continuous for all values of $t$
Therefore, the domain of $tan^{-1}~(1+e^{-t^2})$ is $(-\infty, \infty)$ and this function is continuous at every number in its domain.
